Dynamic learning and strategic communication

Abstract

This paper investigates a dynamic model of strategic communication between a principal and an expert with conflicting preferences. In each period, the uninformed principal selects an experiment which privately reveals information about an unknown state to the expert. The expert then sends a cheap talk message to the principal. We show that the principal can elicit perfect information from the expert about the state and achieve the first-best outcome in only two periods if the expert’s preference bias is not too large. If the state space is unbounded, full information revelation is possible for an arbitrarily large bias. Moreover, full revelation of information is feasible in more general frameworks than those considered in the literature, including frameworks which include non-quasiconcave and non-supermodular payoff functions and those with a privately known bias of the expert.

Appendix

A perfect Bayesian equilibrium includes the learning protocol $\left\{ I_{1}^{*},I_{2}^{*}\left( m_{1},I_{1}\right) \right\}$, the decision rule $a^{*}\left( .\right)$, the belief $\mu ^{*}\left( .\right)$, and the expert’s strategy $\left\{ \sigma _{1}^{*}\left( .\right) ,\sigma _{2}^{*}\left( .\right) \right\}$, such that $\mu ^{*}\left( .\right)$ is consistent with the players’ strategies and the strategies satisfy the following conditions.

• (1) Given any $\left\{ I_{1},I_{2},m_{1},m_{2}\right\}$ and $\mu _{2}^{*}\left( .\right) , a^{*}$ maximizes the principal’s payoff:

  $$\begin{aligned} \underset{y\in \mathbb {R}}{a^{*}\in \arg \max }\, E_{\theta }\left[ U\left( y,\theta \right) |\mu _{2}^{*}\left( m_{1},m_{2},I_{1},I_{2}\right) \right] . \end{aligned}$$

• (2) Given $a^{*}\left( .\right)$ and any $\left\{ I_{1} ,I_{2},s_{1},s_{2},m_{1}\right\} , \sigma _{2}^{*}$ maximizes the expert’s second-period payoff:²⁵

  $$\begin{aligned} m_{2}\underset{x\in \mathcal {M}}{\in \arg \max }E_{\theta }\left[ V\left( a^{*}\left( m_{1},x,I_{1},I_{2}\right) ,\theta ,b\right) |I_{1} ,I_{2},s_{1},s_{2}\right] . \end{aligned}$$

• (3) Given $\left\{ \sigma _{1}^{*}\left( .\right) ,\sigma _{2}^{*}\left( .\right) ,a^{*}\left( .\right) \right\} ,\mu _{1}^{*}\left( .\right)$, and any $\left\{ m_{1},I_{1}\right\}$, the information structure $I_{2}^{*}\left( m_{1},I_{1}\right)$ maximizes the principal’s interim payoff:

  $$\begin{aligned} I_{2}^{*}\underset{I_{2}\in \mathcal {I}}{\in \arg \max }E_{\theta ,s_{1},s_{2} }\left[ U\left( a^{*}\left( m_{1},\sigma _{2}^{*}\left( I_{1} ,I_{2},s_{1},s_{2},m_{1}\right) ,I_{1},I_{2}\right) ,\theta \right) |I_{1},I_{2},m_{1}\right] . \end{aligned}$$

• (4) Given $\left\{ a^{*}\left( .\right) ,\sigma _{2}^{*}\left( .\right) ,I_{2}^{*}\left( .\right) \right\}$, and any $\left\{ I_{1},s_{1}\right\} , \sigma _{1}^{*}$ maximizes the expert’s interim payoff:

  $$\begin{aligned}&m_{1}\underset{x\in \mathcal {M}}{\in \arg \max }E_{\theta ,s_{2}}\left[ V\left( a^{*}\left( x,\sigma _{2}^{*}\left( I_{1},I_{2}\left( x,I_{1}\right) ,s_{1},s_{2},x\right) ,I_{1},I_{2}\left( x,I_{1}\right) \right) , \right. \right. \\&\quad \left. \left. \theta ,b\right) |I_{1},I_{2}\left( x,I_{1}\right) ,s_{1}\right] . \end{aligned}$$

• (5) Given $\left\{ \sigma _{1}^{*}\left( .\right) ,\sigma _{2}^{*}\left( .\right) ,a^{*}\left( .\right) ,I_{2}^{*}\left( .\right) \right\}$, the information structure $I_{1}^{*}\in \mathcal {I}\$maximizes the principal’s ex-ante payoff.

Now, we derive the expressions for $p_{s}$ and $p_{s}^{c}$. Split $\Theta$ into two subintervals, $\left[ 0,\bar{s}\right]$ and $\left[ \bar{s},1\right]$. Then, split the subinterval $\left[ 0,\bar{s}\right]$ into $N$ subintervals of the equal length $\Delta =\frac{\bar{s}}{N}$, and consider two-interval sets $S_{s,\Delta }=S_{s,\Delta }^{1}\cup S_{s,\Delta } ^{2}=\left[ s,s+\Delta \right] \cup \left[ \varphi \left( s\right) ,\varphi \left( s+\Delta \right) \right]$, where $s=0,\frac{\bar{s}}{N},...,\bar{s}-\frac{\bar{s}}{N},S_{s,\Delta }^{1}=\left[ s,s+\Delta \right]$, and $S_{s,\Delta }^{2}=\left[ \varphi \left( s\right) ,\varphi \left( s+\Delta \right) \right]$. Then, we have

$$\begin{aligned} \ p_{s}\left( \Delta \right) =\frac{\Pr [\theta \in S_{s,\Delta }^{1}]}{\Pr \left[ \theta \in S_{s,\Delta }\right] }\!=\!\frac{\frac{1}{\Delta }\left( F\left( s\!+\!\Delta \right) \!-\!F\left( s\right) \right) }{\frac{1}{\Delta }\left( F\left( s+\Delta \right) \!-\!F\left( s\right) \right) \! +\!\frac{1}{\Delta }\left( F\left( \varphi \left( s+\Delta \right) \right) \!-\!F\left( \varphi \left( s\right) \right) \right) }, \end{aligned}$$

which gives $p_{s}=\underset{\Delta \rightarrow 0}{\lim }p_{s}\left( \Delta \right) =\frac{f\left( s\right) }{f\left( s\right) +f\left( \varphi \left( s\right) \right) \varphi ^{\prime }\left( s\right) }$ and $p_{s}^{c}=\underset{\Delta \rightarrow 0}{\lim }\left( 1-p_{s}\left( \Delta \right) \right) =1-p_{s}$. Since $f\left( s\right) >0,f\left( \varphi \left( s\right) \right) >0$, and $\varphi ^{\prime }\left( s\right) >0$ for all $s\in \left[ 0,\bar{s}\right]$, then $p_{s}>0$ and $p_{s}^{c} >0$.

Proof of Theorem 1

Consider the principal’s decision rule defined by (4). Upon observing $s=1$, the expert prefers to reveal it truthfully in both periods, since $V\left( a_{p}\left( 1\right) ,1,b\right) \ge V\left( a,1,b\right) ,a\in \mathcal {A}$. After observing a signal $s\in [0,\bar{s})$ in the first period, the expert’s optimal strategy is either to learn $\theta$ in the second period by reporting $s$ truthfully and inducing an action $a\in \left\{ a_{p}\left( s\right) ,a_{p}\left( \varphi \left( s\right) \right) \right\}$ or to induce the optimal interim decision $y_{1}\left( s,b\right) \in Y_{1}\left( s,b\right)$.

Consider the expert’s second-period incentive-compatibility constraints (7)–(8) given truthful reporting in $t=1$. If $b=0$, then $a_{p}\left( s\right) =y\left( s,0\right) ,\forall s$. Since $a_{p}\left( s\right) \ne a_{p}\left( \varphi \left( s\right) \right) ,s\in \left[ 0,\bar{s}\right]$, then

$$\begin{aligned}&V\left( a_{p}\left( s\right) ,s,0\right) -V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,0\right) =V\left( y\left( s,0\right) ,s,0\right) -V\left( y\left( \varphi \left( s\right) ,0\right) ,s,0\right) >0, \\&s\in \left[ 0,\bar{s}\right] , \text {and}\\&\quad V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,0\right) -V\left( a_{p}\left( s\right) ,\varphi \left( s\right) ,0\right) >0, \quad s\in \left[ 0,\bar{s}\right] . \end{aligned}$$

By the continuity of $V\left( a,s,b\right) ,a_{p}\left( s\right)$, and $\varphi \left( s\right)$ in $\left( a,s,b\right)$ and the compactness of $\left[ 0,\bar{s}\right]$, there are $b_{s}>0$ and $b_{\varphi }>0$, such that²⁶

$$\begin{aligned} V\left( a_{p}\left( s\right) ,s,b\right)&\ge V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,b\right) ,s\in \left[ 0,\bar{s}\right] , \quad b\le b_{s},\text {and}\\ V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right)&\ge V\left( a_{p}\left( s\right) ,\varphi \left( s\right) ,b\right) ,s\in \left[ 0,\bar{s}\right] , \quad b\le b_{\varphi }. \end{aligned}$$

The expert’s interim payoffs $E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right]$ and $E\left[ V\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right]$ from truthtelling in both periods and inducing $y_{1}\left( s,b\right)$, respectively, are

$$\begin{aligned}&E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] =p_{s}V\left( a_{p}\left( s\right) ,s,b\right) +\left( 1-p_{s}\right) V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) ,\\&\quad \text {and} \; E\left[ V\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right] =\underset{a\in \mathcal {A}}{\max }\, \, p_{s}V\left( a,s,b\right) +\left( 1-p_{s}\right) V\left( a,\varphi \left( s\right) ,b\right) . \end{aligned}$$

Suppose $b=0$. Then, $y\left( \varphi \left( s\right) ,0\right) =a_{p}\left( \varphi \left( s\right) \right) \ne a_{p}\left( s\right) =y\left( s,0\right)$ implies $y_{1}\left( s,0\right) \ne y\left( s,0\right)$ and/or $y_{1}\left( s,0\right) \ne y\left( \varphi \left( s\right) ,0\right)$. Because $0<p_{s}<1$, we have

$$\begin{aligned} E\left[ V\left( y_{1}\left( s,0\right) ,\theta ,0\right) |s\right]&=p_{s}V\left( y_{1}\left( s,0\right) ,s,0\right) +\left( 1-p_{s}\right) V\left( y_{1}\left( s,0\right) ,\varphi \left( s\right) ,0\right) \\&<p_{s}V\left( y\left( s,0\right) ,s,0\right) +\left( 1-p_{s}\right) V\left( y\left( \varphi \left( s\right) ,0\right) ,\varphi \left( s\right) ,0\right) \\&=p_{s}V\left( a_{p}\left( s\right) ,s,0\right) +\left( 1-p_{s}\right) V\left( a_{p}\left( \varphi \left( s\right) ,0\right) ,\varphi \left( s\right) ,0\right) \\&=E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,0\right) |s\right] , \quad s\in \left[ 0,\bar{s}\right] . \end{aligned}$$

Because $\mathcal {A}$ is an image of a continuous function $a_{p}\left( \theta \right)$ on the compact $\Theta$, it is compact (Arkhangel’skii and Pontrjagin 2011). By the continuity of $\varphi \left( s\right) ,p_{s},V\left( a,s,b\right)$, and $a_{p}\left( s\right)$ in $\left( a,s,b\right)$, it follows that $E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right]$ and $E\left[ V\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right]$ are continuous in $\left( s,b\right)$ (Sydsæter et al. 2005). Hence, there is $b_{y}>0$ such that

$$\begin{aligned} E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] \ge E\left[ V\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right] , \quad s\in \left[ 0,\bar{s}\right] , \; b\le b_{y}. \end{aligned}$$

Thus, there is the fully informative equilibrium if $b\le \bar{b}$, where $\bar{b}=\min \left\{ b_{s},b_{\varphi },b_{y}\right\} >0$. $\square$

Proof of Lemma 1

Suppose there is the fully informative equilibrium for $\bar{b}>0$, that is,

$$\begin{aligned} V\left( a_{p}\left( s\right) ,s,\bar{b}\right) -V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,\bar{b}\right) \ge 0, \end{aligned}$$

(11)

and

$$\begin{aligned} \Delta V\left( y_{1}\left( s,\bar{b}\right) ,s,\bar{b}\right)= & {} E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,\bar{b}\right) |s\right] -E\left[ V\left( y_{1}\left( s,\bar{b}\right) ,\theta ,\bar{b}\right) |s\right] \\= & {} p_{s}V\left( a_{p}\left( s\right) ,s,\bar{b}\right) +\left( 1-p_{s}\right) V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,\bar{b}\right) \\&-p_{s}V\left( y_{1}\left( s,\bar{b}\right) ,s,\bar{b}\right) \!-\!\left( 1\!-\!p_{s}\right) V\left( y_{1}\left( s,\bar{b}\right) ,\varphi \left( s\right) ,\bar{b}\right) \ge 0, \end{aligned}$$

for all $s\in [0,1/2)$. We will prove that there exists the fully informative equilibrium for any $b<\bar{b}$.

Because $a_{p}\left( \varphi \left( s\right) \right) >a_{p}\left( s\right)$ and $V_{ab}^{\prime \prime }>0$, then $V_{b}^{\prime }\left( a_{p}\left( s\right) ,s,b\right) -V_{b}^{\prime }\left( a_{p}\left( \varphi \left( s\right) \right) ,s,b\right) <0,s\in [0,1/2)$. This implies that (11) holds for $b<\bar{b}$:

$$\begin{aligned}&V\left( a_{p}\left( s\right) ,s,b\right) -V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,b\right) >V\left( a_{p}\left( s\right) ,s,\bar{b}\right) \nonumber \\&\quad -V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,\bar{b}\right) \ge 0, \quad b<\bar{b}. \end{aligned}$$

(12)

Also, $a_{p}\left( \varphi \left( s\right) \right) >y\left( s,b\right) ,s\in [0,1/2),b<\bar{b}$. $\big ({\text {Otherwise, if}} \; a_{p}\left( \varphi \left( s\right) \right) \le y\left( s,b\right) ,$ then $a_{p}\left( s\right) <a_{p}\left( \varphi \left( s\right) \right) \le y\left( s,b\right) \; \mathrm{{and}} \; V_{a}^{\prime }\left( a,\theta ,b\right) >0,a<y\left( \theta ,b\right)$ result in $V\left( a_{p}\left( s\right) , s, b\right) <V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,b\right) \big )$.

Since $V_{ab}^{\prime \prime }>0$, then $V_{b}^{\prime }\left( a,s,b\right)$ is increasing in $a$. Because $a_{p}\left( s\right) <y\left( s,b\right) ,\forall s,b>0$, we have:

$$\begin{aligned}&V_{b}^{\prime }\left( a_{p}\left( s\right) ,s,b\right) <V_{b}^{\prime }\left( y\left( s,b\right) ,s,b\right) \le 0, \text {and}E\left[ V_{b}^{\prime }\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] \\&\quad =p_{s}V_{b}^{\prime }\left( a_{p}\left( s\right) ,s,b\right) +\left( 1-p_{s}\right) V_{b}^{\prime }\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) <0. \end{aligned}$$

Because $y_{1}\left( s,b\right) >a_{p}\left( 0\right)$, we need to consider two cases: $y_{1}\left( s,b\right) <a_{p}\left( 1\right)$ and $y_{1}\left( s,b\right) =a_{p}\left( 1\right)$. First, let $y_{1}\left( s,b\right) <a_{p}\left( 1\right)$. In this case, $y_{1}\left( s,b\right) =y_{1}^{*}\left( s,b\right)$ and $\Delta V\left( y_{1}\left( s,b\right) ,s,b\right) =\Delta V\left( y_{1}^{*}\left( s,b\right) ,s,b\right)$. This implies $\Delta V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,s,b\right) =\Delta V_{b}^{\prime }\left( y_{1}^{*}\left( s,b\right) ,s,b\right) \le 0$. Second, let $y_{1}\left( s,b\right) =a_{p}\left( 1\right)$, which means $y_{1}\left( s,b\right) >a_{p}\left( \varphi \left( s\right) \right) >a_{p}\left( s\right)$. Since $V_{ab}^{\prime \prime }>0$, then $V_{b}^{\prime }\left( a_{p}\left( s\right) ,s,b\right) <V_{b}^{\prime }\left( a_{p}\left( 1\right) ,s,b\right)$ and $V_{b}^{\prime }\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) <V_{b}^{\prime }\left( a_{p}\left( 1\right) ,\varphi \left( s\right) ,b\right)$. By the Envelope theorem, $\frac{d}{db}E\left[ V\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right] =E\left[ V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right]$ results in:

$$\begin{aligned} \frac{d\Delta V\left( y_{1}\left( s,b\right) ,s,b\right) }{db}= & {} p_{s} V_{b}^{\prime }\left( a_{p}\left( s\right) ,s,b\right) +\left( 1-p_{s}\right) V_{b}^{\prime }\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) \\&-p_{s}V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,s,b\right) -\left( 1-p_{s}\right) V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,\varphi \left( s\right) ,b\right) \\= & {} p_{s}\left( V_{b}^{\prime }\left( a_{p}\left( s\right) ,s,b\right) -V_{b}^{\prime }\left( a_{p}\left( 1\right) ,s,b\right) \right) +\left( 1-p_{s}\right) \\&\times \left( V_{b}^{\prime }\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) -V_{b}^{\prime }\left( a_{p}\left( 1\right) ,\varphi \left( s\right) ,b\right) \right) <0. \end{aligned}$$

Thus, $\Delta V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,s,b\right) \le 0$ and $\Delta V\left( y_{1}\left( s,b\right) ,s,b\right) \ge \Delta V\left( y_{1}\left( s,\bar{b}\right) ,s,\bar{b}\right) \ge 0,b<\bar{b}$. $\square$

Proof of Corollary 1

The decision $y_{1}^{*}\left( s,b\right)$ is given by the first-order condition:

$$\begin{aligned} p_{s}V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,s\right) ,s\right) +\left( 1-p_{s}\right) V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,\varphi \left( s\right) \right) ,\varphi \left( s\right) \right) =0, \end{aligned}$$

where $V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,s\right) ,s\right) <0<V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,\varphi \left( s\right) \right) ,\varphi \left( s\right) \right)$ due to $V_{a\theta }^{\prime \prime }>0$. Because $V_{b}^{\prime }\left( a-\beta \left( b,\theta \right) ,\theta \right) =-V_{a}^{\prime }\left( a-\beta \left( b,\theta \right) ,\theta \right) \beta _{b}^{\prime }\left( b,\theta \right)$ and $\beta _{b}^{\prime }\left( b,s\right) \ge 0$, this implies $-V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,s\right) ,s\right) \beta _{b}^{\prime }\left( b,s\right) \ge 0$, and

$$\begin{aligned} E\left[ V_{b}^{\prime }\left( y_{1}^{*}\left( s,b\right) ,\theta ,b\right) |s\right]= & {} -p_{s}V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,s\right) ,s\right) \beta _{b}^{\prime }\left( b,s\right) -\left( 1-p_{s}\right) \\&\times V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,\varphi \left( s\right) \right) ,\varphi \left( s\right) \right) \beta _{b}^{\prime }\left( b,\varphi \left( s\right) \right) \\\ge & {} -\beta _{b}^{\prime }\left( b,\varphi \left( s\right) \right) \left( p_{s}V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,s\right) ,s\right) \right. \\&\left. +\left( 1-p_{s}\right) V_{a}^{\prime }\left( y_{1}^{*}\left( s,b\right) -\beta \left( b,\varphi \left( s\right) \right) ,\varphi \left( s\right) \right) \right) =0, \end{aligned}$$

where the inequality follows from $\beta _{b}^{\prime }\left( b,s\right) \ge \beta _{b}^{\prime }\left( b,\varphi \left( s\right) \right)$ due to $\beta _{b\theta }^{\prime \prime }\left( b,\theta \right) \le 0$. Thus, $E\left[ V_{b}^{\prime }\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] <0$ and $E\left[ V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,\theta ,b\right) |s\right] \ge 0$, which gives

$$\begin{aligned} \Delta V_{b}^{\prime }\left( y_{1}\left( s,b\right) ,s,b\right) =E\left[ V_{b}^{\prime }\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] -E\left[ V_{b}^{\prime }\left( y_{1}^{*}\left( s,b\right) ,\theta ,b\right) |s\right] <0. \end{aligned}$$

$\square$

Proof of Theorem 2

According to (A2), there exists $\delta \left( b\right) >0$ such that $\left| y\left( s,b\right) -a_{p}\right. \left. \left( s\right) \right| \le \delta \left( b\right) ,\forall s$. By (A3), there exists $\bar{\delta }=\bar{\delta }\left( b,\delta \left( b\right) \right) >0$ such that $V\left( y\left( s,b\right) -\delta \left( b\right) ,s,b\right) \ge V\left( y\left( s,b\right) +\bar{\delta },s,b\right)$ and $V\left( y\left( s,b\right) +\delta \left( b\right) ,s,b\right) \ge V\left( y\left( s,b\right) -\bar{\delta },s,b\right) ,\forall s$. Then, $\bar{\delta }\ge \delta \left( b\right)$. (Otherwise, if $\bar{\delta }<\delta \left( b\right)$, then $V\left( y\left( s,b\right) -\delta \left( b\right) ,s,b\right) \ge V\left( y\left( s,b\right) +\bar{\delta },s,b\right) >V\left( y\left( s,b\right) +\delta \left( b\right) ,s,b\right) \ge V\left( y\left( s,b\right) -\bar{\delta },s,b\right)$, where the first and the last inequalities follow from (A3), and the second one follows from $y\left( s,b\right) +\bar{\delta }<y\left( s,b\right) +\delta \left( b\right)$ and $V_{a}^{\prime }\left( a,s,b\right) <0,a>y\left( s,b\right)$ due to the strict pseudo-concavity of $V\left( a,s,b\right)$. By the same property, we have the contradiction $y\left( s,b\right) -\delta \left( b\right) >y\left( s,b\right) -\bar{\delta }$, or $\delta \left( b\right) <\bar{\delta }$.)

Put $\varphi \left( \theta \right) =\theta +\frac{d}{2}$. By (A4), there exists $d=d\left( b,\bar{\delta }\right) >0$, such that

$$\begin{aligned} y\left( s,b\right) +\bar{\delta }<y_{1}^{*}\left( s,\varphi \left( s\right) ,b\right) <y\left( \varphi \left( s\right) ,b\right) -\bar{\delta }, \quad \forall s. \end{aligned}$$

(13)

$\square$

Construct the learning protocol as follows:

$$\begin{aligned} s_{1}\left( \theta \right)&=\left\{ \begin{array}{ll} \theta &{} \quad \text {if} \, \theta \in [kd,kd+\frac{d}{2}),k=0,1,\ldots ,\\ \varphi ^{-1}\left( \theta \right) &{} \quad \text {otherwise, and} \end{array} \right. \\ s_{2}\left( \theta ,m_{1}\right)&= \left\{ \begin{array}{ll} \theta &{} \quad \text {if} \, \theta =m_{1}\; \text {or} \; \theta =\varphi \left( m_{1}\right) , \text {and} \; m_{1}\in [kd,kd+\frac{d}{2}),\\ &{} \quad k=0,1,\ldots ,\\ s_{0} \notin \mathbb {R}_{+} &{}\quad \text {otherwise}. \end{array}\right. \end{aligned}$$

Consider the expert’s second-period incentive-compatibility constraints given truthful reporting of $s$ in the first period and observing $\theta =s<\varphi \left( s\right)$ in the second period. First, we have

$$\begin{aligned} a_{p}\left( \varphi \left( s\right) \right) \ge y\left( \varphi \left( s\right) ,b\right) -\delta \left( b\right) \ge y\left( \varphi \left( s\right) ,b\right) -\bar{\delta }>y\left( s,b\right) +\bar{\delta }, \quad \forall s, \end{aligned}$$

where the first inequality follows from (A2) and the third inequality follows from (13). This gives

$$\begin{aligned} V\left( a_{p}\left( s\right) ,s,b\right) \ge V\left( y\left( s,b\right) +\bar{\delta },s,b\right) >V\left( a_{p}\left( \varphi \left( s\right) \right) ,s,b\right) , \quad \forall s, \end{aligned}$$

(14)

where the first inequality follows from (A3) if $a_{p}\left( s\right) \le y\left( s,b\right)$ and from $V_{a}^{\prime }\left( a,s,b\right) <0,a>y\left( s,b\right)$ if $a_{p}\left( s\right) >y\left( s,b\right)$. This property also implies the second inequality. Similarly, if the expert observes $\theta =\varphi \left( s\right)$ in the second period, then

$$\begin{aligned} a_{p}\left( s\right) \le y\left( s,b\right) +\delta \left( b\right) \le y\left( s,b\right) +\bar{\delta }<y\left( \varphi \left( s\right) ,b\right) -\bar{\delta }, \quad \forall s, \end{aligned}$$

where the first inequality follows from (A2) and the third inequality follows from (13). This gives

$$\begin{aligned} V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) \!\ge \! V\left( y\left( \varphi \left( s\right) ,b\right) \!-\!\bar{\delta },\varphi \left( s\right) ,b\right) \! >\! V\left( a_{p}\left( s\right) ,\varphi \left( s\right) ,b\right) , \quad \forall s, \end{aligned}$$

(15)

where the first inequality follows from (A3) if $a_{p}\left( \varphi \left( s\right) \right) \ge y\left( \varphi \left( s\right) ,b\right)$ and from $V_{a}^{\prime }\left( a,\varphi \left( s\right) ,b\right) >0,a<y\left( \varphi \left( s\right) ,b\right)$ if $a_{p}\left( \varphi \left( s\right) \right) <y\left( \varphi \left( s\right) ,b\right)$. This property also implies the second inequality.

Consider now the expert’s interim payoff in the first period. First, note that

$$\begin{aligned} V\left( a_{p}\left( s\right) ,s,b\right) \ge V\left( y\left( s,b\right) +\bar{\delta },s,b\right) >V\left( y_{1}^{*}\left( s,\varphi \left( s\right) ,b\right) ,s,b\right) , \quad \forall s, \end{aligned}$$

where the first inequality follows from (14), and the second inequality follows from (A4) and $V_{a}^{\prime }\left( a,s,b\right) <0,a>y\left( s,b\right)$. Similarly, we obtain

$$\begin{aligned} V\!\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) \!\ge \! V\!\left( y\left( \varphi \left( s\right) ,b\right) \!-\!\bar{\delta },\varphi \left( s\right) ,b\right) \!>\!\! V\!\left( y_{1}^{*}\!\left( s,\varphi \left( s\right) ,b\right) ,s,b\right) ,\,\, \forall s, \end{aligned}$$

where the first inequality follows from (15), and the second inequality follows from (A4) and $V_{a}^{\prime }\left( a,\varphi \left( s\right) ,b\right) >0,a<y\left( \varphi \left( s\right) ,b\right)$. As a result,

$$\begin{aligned}&V\left( a_{p}\left( s\right) ,s,b\right) +V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) \nonumber \\&\quad >V\left( y_{1}^{*}\left( s,\varphi \left( s\right) ,b\right) ,s,b\right) +V\left( y_{1}^{*}\left( s,\varphi \left( s\right) ,b\right) ,\varphi \left( s\right) ,b\right) . \end{aligned}$$

(16)

Note that (14)–(15) do not depend on $p_{s}$. Also, the optimal decision in the case of deviation from truthtelling in the first period,

$$\begin{aligned} y_{1}\left( s,s+\frac{d}{2},b\right) = \underset{a\in \mathcal {A} }{\arg \max }\, \, p_{s}\left( V\left( y,s,b\right) +\frac{1-p_{s}}{p_{s} }V\left( y,\varphi \left( s\right) ,b\right) \right) , \end{aligned}$$

is continuous in $p_{s}>0$ and $y_{1}\left( s,\varphi \left( s\right) ,b\right) =y_{1}^{*}\left( s,\varphi \left( s\right) ,b\right)$ if $p_{s}=\frac{1}{2}$. Because (16) holds strictly, there exists an $\varepsilon$–neighborhood of $1$ for each $\left( s,b\right)$, such that if it contains $\frac{1-p_{s}}{p_{s}}=\frac{f\left( \varphi \left( s\right) \right) }{f\left( s\right) }$, then

$$\begin{aligned}&E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] \\&=p_{s}\left( V\left( a_{p}\left( s\right) ,s,b\right) +\frac{f\left( \varphi \left( s\right) \right) }{f\left( s\right) }V\left( a_{p}\left( \varphi \left( s\right) \right) ,\varphi \left( s\right) ,b\right) \right) \\&>p_{s}\left( V\left( y_{1}\left( s,\varphi \left( s\right) ,b\right) ,s,b\right) +\frac{f\left( \varphi \left( s\right) \right) }{f\left( s\right) }V\left( y_{1}\left( s,\varphi \left( s\right) ,b\right) ,\varphi \left( s\right) ,b\right) \right) \\&=E\left[ V\left( y_{1}\left( s,\varphi \left( s\right) ,b\right) ,\theta ,b\right) |s\right] . \end{aligned}$$

Proof of Corollary 2

We construct the learning protocol as that in the proof of Theorem 2 with the distance $\frac{d}{2}$ between the posterior states, which is determined below. Given a signal $s\in [kd,kd+\frac{d}{2}),k\in \mathbb {N}_{0}$ in the first period, truthtelling in both periods provides the payoff to the expert:

$$\begin{aligned} E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right] =-p_{s}\beta ^{2}\left( b,s\right) -p_{s}^{c}\beta ^{2}\left( b,s+\frac{d}{2}\right) \ge -\delta ^{2}. \end{aligned}$$

Inducing the optimal interim decision $y_{1}\left( s,s+\frac{d}{2},b\right) =p_{s}\left( s+\beta \left( b,s\right) \right) +p_{s}^{c}\left( s+\frac{d}{2}+\beta \left( b,s+\frac{d}{2}\right) \right)$ provides the payoff:

$$\begin{aligned} E\left[ V\left( y_{1}\left( s,s+\frac{d}{2},b\right) ,\theta ,b\right) |s\right] =-p_{s}p_{s}^{c}\left( \frac{d}{2}+\beta \left( b,s+\frac{d}{2}\right) -\beta \left( b,s\right) \right) ^{2}. \end{aligned}$$

Because $a_{p}\left( s+\frac{d}{2}\right) -a_{p}\left( s\right) =s+\frac{d}{2}-s=\frac{d}{2}$ and $V\left( a,s,b\right)$ is symmetric in $a$ around $y\left( s,b\right) =s+\beta \left( s,b\right)$, the second-period incentive compatibility constraints are given by:

$$\begin{aligned} a_{p}\left( s+\frac{d}{2}\right) -a_{p}\left( s\right) =\frac{d}{2} \ge 2\beta \left( b,s\right) . \end{aligned}$$

(17)

Since $\beta \left( b,s\right) \le \delta$, (17) are satisfied if $d\ge 4\delta$. Also, (17) implies $\frac{d}{2}-\beta \left( b,s\right) \ge \beta \left( b,s\right) >0$ and

$$\begin{aligned} E\left[ V\left( y_{1}\left( s,s+\frac{d}{2},b\right) ,\theta ,b\right) |s\right]< & {} -p_{s}p_{s}^{c}\left( \frac{d}{2}-\delta \right) ^{2}\\= & {} -\frac{f\left( s\right) f\left( s+\frac{d}{2}\right) }{\left( f\left( s\right) +f\left( s+\frac{d}{2}\right) \right) ^{2}}\left( \frac{d}{2}-\delta \right) ^{2}. \end{aligned}$$

Suppose $f\left( s\right) \ge f\left( s+\frac{d}{2}\right)$. (The case of $f\left( s\right) <f\left( s+\frac{d}{2}\right)$ is symmetric.) If $f\left( \theta \right)$ satisfies condition (A5) for some $\alpha >0$ and $d>0$, then

$$\begin{aligned}&\frac{f\left( s\right) f\left( s+\frac{d}{2}\right) }{\left( f\left( s\right) +f\left( s+\frac{d}{2}\right) \right) ^{2}}=\frac{f\left( s+\frac{d}{2}\right) }{f\left( s\right) }\frac{1}{\left( 1+\frac{f\left( s+\frac{d}{2}\right) }{f\left( s\right) }\right) ^{2}}\ge \frac{e^{-\alpha \frac{d}{2}}}{\left( 1+\frac{f\left( s+\frac{d}{2}\right) }{f\left( s\right) }\right) ^{2}}\ge \frac{e^{-\alpha \frac{d}{2}}}{4}, \\&\quad \text {and} \; E\left[ V\left( y_{1}\left( s,s+\frac{d}{2},b\right) ,\theta ,b\right) |s\right] <-\frac{e^{-\alpha \frac{d}{2}}}{4}\left( \frac{d}{2} -\delta \right) ^{2}=\phi \left( d,\alpha \right) , \end{aligned}$$

where $\phi \left( d,\alpha \right)$ attains the minimum $\phi \left( d^{*}\left( \alpha \right) ,\alpha \right) =-\frac{e^{-2-\alpha \delta } }{\alpha ^{2}}$ at $d^{*}\left( \alpha \right) =\frac{4}{\alpha }+2\delta$ for any $\alpha >0$. Equivalently, each $d>2\delta$ is the minimizer of $\phi \left( x,\alpha \right)$ over $x$ for $\alpha ^{*}\left( d\right) =\frac{4}{d-2\delta }>0$.

Because (A5) holds for $\alpha =\frac{z^{*}}{\delta }$ and $d=d^{*}\left( \alpha \right) =2\delta \left( \frac{2}{z^{*}}+1\right)$, where $z^{*}$ is a unique solution to the equation $z=e^{-1-\frac{z}{2} }$, we have $\alpha \delta =z^{*}=e^{-1-\frac{z^{*}}{2}}=e^{-1-\frac{\alpha \delta }{2}}$ and $e^{-2-\alpha \delta }=\alpha ^{2}\delta ^{2}$. This results in

$$\begin{aligned} E\left[ V\left( a_{p}\left( \theta \right) ,\theta ,b\right) |s\right]\ge & {} -\delta ^{2}=-\frac{e^{-2-\alpha \delta }}{\alpha ^{2}}=\phi \left( d^{*}\left( \alpha \right) ,\alpha \right) \\> & {} E\left[ V\left( y_{1}\left( s,s+\frac{d}{2},b\right) ,\theta ,b\right) |s\right] . \end{aligned}$$

Finally, (17) holds because $z^{*}\simeq 0.314<1$ implies $d=2\delta \left( \frac{2}{z^{*}}+1\right) >4\delta$. $\square$